Uniform Circular Motion

Uniform Circular Motion Physics of motion of a mass in a circle at Constant Speed. Constant Speed  The Magnitude (size) of the velocity vector v is Constant. BUT the DIRECTION of v changes continually! v  r r r v = |v| = constant

Uniform Circular Motion is Circular Motion at Constant Speed. The direction of the velocity is continually changing. The velocity vector is always tangent to the circle.

Acceleration  Time Rate of Change of Velocity Consider details of the motion of a mass m in a circle at Constant Speed. Question: Is there an acceleration? To answer this, consider both Newton’s 1st Law & Newton’s 2nd Law! Recall that by Definition, Acceleration  Time Rate of Change of Velocity That is: a = (Δv/Δt) v = |v| = constant r r

 An object moving in a circle undergoes an acceleration!! Time Rate of Change of Velocity a = (Δv/Δt) Also, recall that both a & v are vectors. Constant Speed  The Magnitude (size) of the velocity vector v is Constant. BUT the DIRECTION of v changes continually!  An object moving in a circle undergoes an acceleration!!

Centripetal (Radial) Acceleration Look at the vector velocity change Δv in the limit that the time interval Δt becomes infinitesimally small & get: Similar Triangles  (Δv/v) ≈ (Δℓ/r) As Δt  0, Δθ  0, A B As Δt  0, Δv   v & Δv is in the radial direction  a  aC is radial!

Δv  0 Close view of the fact that, in the infinitesimal limit, & its direction is towards the center of the circle.

Centripetal Acceleration & sometimes called the Radial Acceleration. This type of acceleration is sometimes called the Centripetal Acceleration & sometimes called the Radial Acceleration. Its vector direction is Radially Inward! The word “Centripetal” is from Greek. It means “Towards the Center” “Centripetal Acceleration”  Center Directed or Center Seeking Acceleration

is always Radially Inward! Below is a typical figure for a particle moving in uniform circular motion, radius r (speed v = constant): The velocity vector v is always tangent to the circle! The centripetal acceleration vector aC is always Radially Inward!  aC  v always!!

Period T  The time for one revolution Period & Frequency Consider again a particle moving in uniform circular motion of radius r (speed v = constant). One common way to describe this motion is in terms of the Period T & the frequency f. Period T  The time for one revolution (time to go around once!) Frequency f  The # of revolutions per second. They are obviously related by: T = (1/f)

aC = (v2/r) = (4π2r/T2) Circumference = Distance Around= 2πr Consider again a particle moving in uniform circular motion, radius r (speed v = constant) Circumference = Distance Around= 2πr  Speed: v = (2πr/T) = 2πrf  Centripetal Acceleration: aC = (v2/r) = (4π2r/T2)

Example: Acceleration of a Revolving Ball A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes f = 2.0 revs/second (period T = 0.5 s). Calculate its centripetal acceleration. Figure 5-10. Caption: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path. Answer: a = v2/r; we can find v from the radius and frequency. v = 7.54 m/s, so a = 94.7 m/s2. r r

Example: Acceleration of a Revolving Ball A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes f = 2.0 revs/second (period T = 0.5 s). Calculate its centripetal acceleration. Solution v = (2πr/T) = 7.54 m/s Figure 5-10. Caption: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path. Answer: a = v2/r; we can find v from the radius and frequency. v = 7.54 m/s, so a = 94.7 m/s2. r r

Example: Acceleration of a Revolving Ball A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes f = 2.0 revs/second (period T = 0.5 s). Calculate its centripetal acceleration. Solution v = (2πr/T) = 7.54 m/s aC = (v2/r) = 94.7 m/s2 Figure 5-10. Caption: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path. Answer: a = v2/r; we can find v from the radius and frequency. v = 7.54 m/s, so a = 94.7 m/s2. r r

Example: Moon’s Centripetal Acceleration The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km (3.84  108 m) and a period T of 27.3 days (2.36  106 s). Calculate the acceleration of the Moon toward the Earth. Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g.

Example: Moon’s Centripetal v = (2πr/T), aC = (v2/r) = (4π2r/T2) Acceleration The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km (3.84  108 m) & a period T of 27.3 days (2.36  106 s). Calculate the acceleration of the Moon toward the Earth. Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g. Solution v = (2πr/T), aC = (v2/r) = (4π2r/T2)

Solution v = (2πr/T), aC = (v2/r) = (4π2r/T2) aC = 2.72  10-3 m/s2 Example: Moon’s Centripetal Acceleration The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km (3.84  108 m) and a period T of 27.3 days (2.36  106 s). Calculate the acceleration of the Moon toward the Earth. Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g. Solution v = (2πr/T), aC = (v2/r) = (4π2r/T2) aC = 2.72  10-3 m/s2

Examples 5.1 & 5.2: A Bug on a CD Calculate A bug sits on the edge of a CD, of radius r = 6 cm (0.06m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g. a. The period of the motion. b. The speed of the bug. c. The centripetal acceleration of the bug.

Because f = 6 rev/s, T = 1/f = 0.17s Examples 5.1 & 5.2: A Bug on a CD A bug sits on the edge of a CD, of radius r = 6 cm (0.06m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g. a. The period of the motion. Because f = 6 rev/s, T = 1/f = 0.17s b. The speed of the bug. c. The centripetal acceleration of the bug.

Because f = 6 rev/s, T = 1/f = 0.17s Examples 5.1 & 5.2: A Bug on a CD A bug sits on the edge of a CD, of radius r = 6 cm (0.06m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g. a. The period of the motion. Because f = 6 rev/s, T = 1/f = 0.17s b. The speed of the bug. v = (2πr/T) = 2.3 m/s c. The centripetal acceleration of the bug.

Because f = 6 rev/s, T = 1/f = 0.17s Examples 5.1 & 5.2: A Bug on a CD A bug sits on the edge of a CD, of radius r = 6 cm (0.06m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g. a. The period of the motion. Because f = 6 rev/s, T = 1/f = 0.17s b. The speed of the bug. v = (2πr/T) = 2.3 m/s c. The centripetal acceleration of the bug. aC = (v2/r) = 88.2 m/s2

Newton’s Laws + Circular Motion

It’s straightforward to see how Newton’s 2nd Law can be applied to circular motion: Since the acceleration is directed toward the center of the circle, the net force must be in that direction also! This “Centripetal Force” can be supplied by a variety of physical objects or forces Also, the “circle” does not need to be a complete circle.

Uniform Circular Motion; Dynamics Consider a particle moving in uniform circular motion at radius r & speed v = constant. Centripetal Acceleration is: aC = (v2/r) , aC  v always!! aC is radially inward always! Newton’s 1st Law  There must be a force acting! Newton’s 2nd Law  ∑F = ma = maC = m(v2/r) Direction: The total force must be radially inward always!

A force is required to keep an object moving in a circle A force is required to keep an object moving in a circle. If the speed is constant, the force is directed toward the center of the circle. The direction of the force is continually changing so that it is always pointed toward the center of the circle. ∑F = ma = maC = m(v2/r) Example: A ball twirled on the end of a string. In that case, the force is the tension in the string.

(It is a center directed force) “Centripetal Force” Newton’s 2nd Law: ∑F = ma = maC= m(v2/r) The total force ∑F must be radially inward always! The force which enters Newton’s 2nd Law in this case is often called a “Centripetal Force”. (It is a center directed force) The “Centripetal Force” is NOT a new kind of force! It could be string tension, gravity, etc. It’s the right side of ∑F = ma, not the left side! It’s the form of “ma”, for circular motion!

the centripetal force must be inward You can understand that the centripetal force must be inward by thinking about the ball on a string. Strings only pull; they never push. MISCONCEPTION!! The force on the ball is NEVER outward (“Centrifugal”). It is ALWAYS inward (Centripetal) !!

ON THE BALL is Inward (centripetal). An outward (“centrifugal”) force ON THE BALL is NOT a valid concept! The string tension force ON THE BALL is Inward (centripetal). F F The string tension force F on the ball is INWARD toward the center of the circle!

What happens if the cord on the ball is broken or released? For the ball to move in a circle, there must be an inward (Centripetal) force pointed towards the circle center so that the natural tendency of the object to move in a straight line (Newton’s 1st Law!) will be overcome. If the centripetal force goes to zero, the ball will fly off in a direction tangent to the circle (Newton’s 1st Law again!) There is no centrifugal force pointing outward on the ball!

∑F = ma  FTx = max= maR = m(v2/r) Example (Estimate) Estimate the force a person must exert on a string attached to a 0.15 kg ball to make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2 rev/s. m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s Assumption: Circular path is  in horizontal plane, so φ  0  cos(φ)  1 Newton’s 2nd Law: ∑F = ma  FTx = max= maR = m(v2/r) v = (2πr/T) = 7.54 m/s So, the tension is (approximately) FTx  14 N Figure 5-17. Answer: Ignoring the weight of the ball, so FT is essentially horizontal, we find FT = 14 N.

Example: Revolving Ball (Vertical Circle) A ball, mass m = 0.15 kg on the end of a (massless) cord of length r = 1.1 m cord is swung in a vertical circle. Calculate: a. The minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. b. The tension in the cord at the bottom of the arc, assuming that there the ball is moving at twice the speed found in part a. Hint: The minimum speed at the top will happen for the minimum tension FT1 Note: Here, string tension & gravity are acting together (both enter Newton’s 2nd Law!) to produce centripetal acceleration. Figure 5-18. Caption: Example 5–12. Freebody diagrams for positions 1 and 2. Solution: See the freebody diagrams. The minimum speed occurs when the tension is zero at the top; v = 3.283 m/s. Substitute to find FT2 = 7.35 N.

Problem FT1 = 3.73 N FT2 = 9.61 N Use: ∑F = maR r = 0.72 m, v = 4 m/s, m = 0.3 kg Use: ∑F = maR At the top of the circle, Newton’s 2nd Law is: (down is positive!) FT1 + mg = m(v2/r) FT1 = 3.73 N At the bottom of the circle: Newton’s 2nd Law is: (up is positive!) FT2 - mg = m(v2/r) FT2 = 9.61 N

Example: Back to a Bug on a CD From Examples 5.1 & 5.2 A bug sits on the edge of a CD, of radius r = 0.06m, as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Results from before: Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10-3 m/s2 = 2.78 x 10-4 g. T = 0.17s, v = 2.3 m/s, aC = 88.2 m/s2 Given: Mass m = 5  10-3 kg Calculate: The force which keeps the bug on the CD F = maC = 0.1 N

Solve for normal force & compare. Conceptual Example A Ferris wheel rider moves in a vertical circle of radius r at constant speed v. Is the normal force FN1 that the seat exerts on the rider at the top of the wheel a. less than, b. more than, or c. equal to the normal force FN2 that the seat exerts on the rider at the bottom of the wheel? Use Newton’s 2nd Law: ∑F = maC at top & bottom. Solve for normal force & compare. 1 2

Conceptual Example  FTx = maC = m(v2/r)  FTy - mg = 0, FTy = mg A tether ball is hit so that it revolves around a pole in a circle of radius r at constant speed v. In what direction is the acceleration? What force causes it? Newton’s 2nd Law: ∑F = ma x: ∑Fx = max  FTx = maC = m(v2/r) y: ∑Fy = may = 0  FTy - mg = 0, FTy = mg

Centripetal Force Example Outside the International Space Station Whirls a ball on a string in a perfect circle. The centripetal acceleration is produced by the tension in the string. If the string breaks, the object would move in a direction tangent to the circle at a constant speed.